Evaluate log\(_2\) 8 — log\(_3\) \(\frac{1}{9}\)
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Correct Answer: Option D
Explanation:
log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\)
= log \(_2\) 2\(^3\) – log\(_3\) 9\(^{-1}\)
= log\(_2\) 2\(^3\) – log\(_3\) 3\(^{-2}\)
Based on law of logarithm
= 3 log\(_2\) 2 – (-2 log\(_3\) 3)
But log\(_2\) 2 = 1,
log\(_3\) 3 = 1
So, = 3 + 2
= 5
log\(_2\) 8 – log\(_3\) \(\frac{1}{9}\)
= log \(_2\) 2\(^3\) – log\(_3\) 9\(^{-1}\)
= log\(_2\) 2\(^3\) – log\(_3\) 3\(^{-2}\)
Based on law of logarithm
= 3 log\(_2\) 2 – (-2 log\(_3\) 3)
But log\(_2\) 2 = 1,
log\(_3\) 3 = 1
So, = 3 + 2
= 5